1s - EE 478 Handout#8 Multiple User Information Theory...

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Unformatted text preview: EE 478 Handout #8 Multiple User Information Theory October 7, 2008 Homework Set #1 Solutions 1. Solution: (a) Consider H ( X | Z ) ≤ H ( X,Y | Z ) = H ( Y | Z ) + H ( X | Y,Z ) ≤ H ( Y | Z ) + H ( X | Y ) . (b) h ( X + Y ) ≥ h ( X ). Since conditioning reduces the differential entropy, h ( X + Y ) ≥ h ( X + Y | Y ) = h ( X ) . (c) I ( g ( X ); Y ) ≤ I ( X ; Y ). I ( g ( X ); Y ) = H ( Y )- H ( Y | g ( X )) ≤ H ( Y )- H ( Y | X ) = I ( X ; Y ) . (d) I ( Y ; Z | X ) ≥ I ( Y ; Z ) if p ( x,y,z ) = p ( x ) p ( y ) p ( z | x,y ). Since X and Y are independent random variables, I ( Y ; Z | X ) = H ( Y | X )- H ( Y | Z,X ) = H ( Y )- H ( Y | Z,X ) ≥ H ( Y )- H ( Y | Z ) = I ( Y ; Z ) . 2. Solution: (a) Consider h ( Y n ) ≤ n X i =1 h ( Y i ) = n X i =1 1 2 log(2 πe ) K ii = 1 2 log(2 πe ) n n Y i =1 K ii . 1 (b) Now since h ( Y n ) = 1 2 log(2 πe ) n | K | , the result in part (a) implies that | K | ≤ n Y i =1 K ii . 3. Solution: The identity can be proved by induction, or more simply using chain rule of mutual information. Notice that by chain rule the mutual information between Z n and W can be expanded in n ! ways, depending on how we order the element of Z n . For instance, these are both valid expansions I ( Z n ; W ) = n X i =1 I ( Z i ; W | Z i- 1 ) = n X j =1 I ( Z j ; W | Z n j +1 ) , where the two ordering Z n = ( Z 1 ,...,Z n ) and Z n = ( Z n ,...,Z 1 ) have been used. Therefore, we have n X i =1 I ( X n i +1 ; Y i | Y i- 1 ) = n X i =1 n X j = i +1 I ( X j ; Y i | Y i- 1 ,X n j +1 ) (1) = n X j =2 j- 1 X i =1 I ( X j ; Y i | Y i- 1 ,X n j +1 ) (2) = n X j =2 I ( X j ; Y j- 1 | X n j +1 ) (3) = n X j =1 I ( X j ; Y j- 1 | X n j +1 ) (4) = n X i =1 I ( Y i- 1 ; X i | X n i +1 ) where (1) and (3) follow from chain rule of mutual information, (2) is obtained switch- ing the order in the summations, and finally (4) follows from the fact that Y = ∅ ....
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1s - EE 478 Handout#8 Multiple User Information Theory...

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